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(C) Given $g(x) = x - [x] = \{x\}$,which represents the fractional part of $x$.Since $f(x)$ is continuous and $f(0) = f(1)$,the composite function $h(

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is continuous on $R$

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(C) Given $g(x) = x - [x] = \{x\}$,which represents the fractional part of $x$.Since $f(x)$ is continuous and $f(0) = f(1)$,the composite function $h(x) = f(\{x\})$ is defined for all $x \in R$.For any integer $n$,as $x 	o n^+$,$g(x) 	o 0^+$,so $h(x) 	o f(0)$.As $x 	o n^-$,$g(x) 	o 1^-$,so $h(x) 	o f(1)$.Since $f(0) = f(1)$,the left-hand limit and right-hand limit at any integer $n$ are equal to $f(0)$,which is also the value $h(n) = f(0)$.Thus,$h(x)$ is continuous at all integers $n$,and since it is a composition of continuous functions on intervals $(n, n+1)$,it is continuous on $R$.

Let $[x]$ denote the integral part of $x \in R$. Let $g(x) = x - [x]$. Let $f(x)$ be any continuous function with $f(0) = f(1)$. Then the function $h(x) = f(g(x))$:

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